Using Moles to Balance Equations (Higher Tier)

This lesson covers how to use experimental mass data and mole calculations to deduce the balanced equation for a reaction, as required by the AQA GCSE Combined Science Trilogy specification (8464) at Higher tier.

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Why Use Moles to Balance Equations?

Sometimes you are given experimental data — masses of reactants and products — and asked to work out the balanced equation. Instead of guessing coefficients, you can calculate the moles of each substance and find the simplest whole-number ratio.

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The Method

flowchart TD
    A["1. Write the correct formulae\nof all reactants and products"] --> B["2. Find the mass of\neach substance"]
    B --> C["3. Calculate the moles of\neach substance using\nmoles = mass ÷ Mr"]
    C --> D["4. Find the simplest\nwhole-number ratio"]
    D --> E["5. Write the balanced\nequation using this ratio"]
    style A fill:#3b82f6,color:#fff,stroke:#2563eb
    style D fill:#f59e0b,color:#000,stroke:#d97706
    style E fill:#22c55e,color:#fff,stroke:#16a34a

Step-by-Step

1. Write the correct formulae of reactants and products.
2. Record the mass of each substance (from the data given).
3. Calculate the moles of each: $\text{moles} = \dfrac{\text{mass}}{M_r}$moles=Mr​mass​
4. Divide all mole values by the smallest to find the simplest ratio.
5. If necessary, multiply up to get whole numbers.

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Worked Examples

Example 1 — Magnesium and Oxygen

Data: 4.8 g of magnesium reacts with 3.2 g of oxygen to form 8.0 g of magnesium oxide.

Substance	Formula	Mass (g)	$M_r$Mr​	Moles
Magnesium	Mg	4.8	24	$\frac{4.8}{24} = 0.2$244.8​=0.2
Oxygen	$\text{O}_2$O2​	3.2	32	$\frac{3.2}{32} = 0.1$323.2​=0.1
Magnesium oxide	MgO	8.0	40	$\frac{8.0}{40} = 0.2$408.0​=0.2

Simplest ratio (divide by smallest, 0.1):

$\text{Mg : O}_2 \text{ : MgO} = \frac{0.2}{0.1} : \frac{0.1}{0.1} : \frac{0.2}{0.1} = 2 : 1 : 2$Mg : O2​ : MgO=0.10.2​:0.10.1​:0.10.2​=2:1:2

$\boxed{2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}}$2Mg+O2​→2MgO​

Example 2 — Iron and Sulfur

Data: 11.2 g of iron reacts with 6.4 g of sulfur to form 17.6 g of iron sulfide.

Substance	Formula	Mass (g)	$M_r$Mr​ or $A_r$Ar​	Moles
Iron	Fe	11.2	56	$\frac{11.2}{56} = 0.2$5611.2​=0.2
Sulfur	S	6.4	32	$\frac{6.4}{32} = 0.2$326.4​=0.2
Iron sulfide	FeS	17.6	88	$\frac{17.6}{88} = 0.2$8817.6​=0.2

Ratio: $0.2 : 0.2 : 0.2 = 1 : 1 : 1$0.2:0.2:0.2=1:1:1

$\boxed{\text{Fe} + \text{S} \rightarrow \text{FeS}}$Fe+S→FeS​

Example 3 — Copper Oxide and Carbon

Data: 8.0 g of copper oxide reacts with 0.6 g of carbon to form 6.4 g of copper and 2.2 g of carbon dioxide.

Substance	Formula	Mass (g)	$M_r$Mr​	Moles
Copper oxide	CuO	8.0	79.5	$\frac{8.0}{79.5} = 0.1006$79.58.0​=0.1006
Carbon	C	0.6	12	$\frac{0.6}{12} = 0.05$120.6​=0.05
Copper	Cu	6.4	63.5	$\frac{6.4}{63.5} = 0.1008$63.56.4​=0.1008
Carbon dioxide	$\text{CO}_2$CO2​	2.2	44	$\frac{2.2}{44} = 0.05$442.2​=0.05

Divide by smallest (0.05):

$\text{CuO : C : Cu : CO}_2 = 2 : 1 : 2 : 1$CuO : C : Cu : CO2​=2:1:2:1

$\boxed{2\text{CuO} + \text{C} \rightarrow 2\text{Cu} + \text{CO}_2}$2CuO+C→2Cu+CO2​​

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Dealing With Non-Whole-Number Ratios

Sometimes dividing by the smallest gives values like 1.5 or 2.5. In these cases, multiply all values by the smallest whole number that gives integers.

Ratio after dividing	Multiply by	Result
1 : 1.5	2	2 : 3
1 : 2.5	2	2 : 5
1 : 1.33	3	3 : 4

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Common Mistakes to Avoid

Mistake	Correction
Using masses directly as the ratio	Masses do not give the correct ratio — you must convert to moles first
Not dividing by the smallest number of moles	This step converts to the simplest ratio
Rounding too early	Keep at least 3 significant figures until the final step
Forgetting to check conservation of mass	Total mass of reactants should equal total mass of products

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Why This Matters

This technique is how chemists discover equations from experimental data. It connects the practical world of weighing and measuring to the theoretical world of balanced equations and mole ratios.

Exam Tip (AQA 8464): If the question gives you masses and asks for a balanced equation, this method is what they want. Show all your working — calculate moles, find the ratio, then write the equation.

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Summary

• Calculate the moles of each substance from the mass data.
• Divide all mole values by the smallest to find the simplest ratio.
• Multiply up if needed to get whole numbers.
• Use the ratio as the coefficients in the balanced equation.
• This is a Higher tier topic for AQA Combined Science Trilogy (8464).

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Extended Worked Examples — Deducing Equations from Mass Data

This deductive skill is a synoptic skill: it combines $M_r$Mr​, moles, conservation of mass and balanced equations. It is a strong discriminator at grade 7–9 on the Higher tier.

Worked Example 4 — Aluminium and Bromine

Data: 5.4 g of aluminium reacts with 48.0 g of bromine to form 53.4 g of aluminium bromide. Deduce the balanced equation.

Substance	Formula	Mass (g)	$M_r$Mr​ or $A_r$Ar​	Moles
Aluminium	Al	5.4	27	$5.4 / 27 = 0.20$5.4/27=0.20
Bromine	$\text{Br}_2$Br2​	48.0	160	$48.0 / 160 = 0.30$48.0/160=0.30
Aluminium bromide	$\text{AlBr}_3$AlBr3​	53.4	267	$53.4 / 267 = 0.20$53.4/267=0.20

Divide by smallest (0.20):

$\text{Al : Br}_2 \text{ : AlBr}_3 = 1 : 1.5 : 1$Al : Br2​ : AlBr3​=1:1.5:1

Multiply by 2:

$2 : 3 : 2$2:3:2

$\boxed{2\text{Al}(s) + 3\text{Br}_2(l) \rightarrow 2\text{AlBr}_3(s)}$2Al(s)+3Br2​(l)→2AlBr3​(s)​

Conservation-of-mass check: $5.4 + 48.0 = 53.4$5.4+48.0=53.4. Balanced and mass-conserved.

Worked Example 5 — Hydrogen and Nitrogen

Data: 2.8 g of $\text{N}_2$N2​ reacts with 0.6 g of $\text{H}_2$H2​ to form 3.4 g of ammonia. Deduce the balanced equation.